/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A world's land speed record was ... [FREE SOLUTION] | 91Ó°ÊÓ

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A world's land speed record was set by Colonel John P. Stapp when in March 1954 he rode a rocket-propelled sled that moved along a track at \(1020 \mathrm{~km} / \mathrm{h}\). He and the sled were brought to a stop in \(1.4 \mathrm{~s}\). (See Fig. \(2-7 .\) ) In terms of \(g\), what acceleration did he experience while stopping?

Short Answer

Expert verified
The acceleration was approximately 20.63 g.

Step by step solution

01

Convert Speed to Meters per Second

The first step is to convert the speed from kilometers per hour to meters per second because standard units for acceleration are in meters and seconds. Given: - The speed is 1020 km/h.To convert km/h to m/s, use the conversion factor: 1 km/h = \(\frac{1}{3.6}\) m/s.Thus, \[1020 \text{ km/h} \times \frac{1}{3.6} = 283.33 \text{ m/s}\]
02

Calculate Deceleration

Now, calculate the deceleration (or negative acceleration) using the formula for acceleration:\[a = \frac{\Delta v}{\Delta t}\]Where:- \(\Delta v\) is the change in velocity.- \(\Delta t\) is the change in time.Given:- \(\Delta v = 283.33 - 0 = 283.33\) m/s (since the final velocity is 0 when stopped)- \(\Delta t = 1.4\) sThus,\[a = \frac{283.33}{1.4} = 202.38 \text{ m/s}^2\]
03

Express Acceleration in Terms of g

To express the deceleration in terms of \( g \), where \( g \approx 9.81 \text{ m/s}^2 \) is the acceleration due to gravity:\[\text{Acceleration in terms of } g = \frac{202.38 \text{ m/s}^2}{9.81 \text{ m/s}^2}\]Calculating gives:\[\frac{202.38}{9.81} \approx 20.63\, g\]
04

Conclusion

Thus, the acceleration experienced by Colonel John P. Stapp while stopping was approximately \( 20.63\, g \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Acceleration
Acceleration is all about how quickly an object changes its velocity. It's not just speeding up; it can be slowing down as well, which is called deceleration. When you hear acceleration, think of it like the "pedal to the metal" in a car, which makes you go faster. However, in physics, it’s about any change in speed or direction.
There are three ways to accelerate:
  • Speeding up (positive acceleration)
  • Slowing down (deceleration or negative acceleration)
  • Changing direction (like turning a corner)
To find acceleration, we use the formula: \[ a = \frac{\Delta v}{\Delta t} \]Where \(\Delta v\) is the change in velocity, and \(\Delta t\) is the time over which this change happens. This formula helps us see exactly how fast your speed is changing over that time. Even stopping, like in the exercise with Colonel John P. Stapp on his rocket sled, involves a form of acceleration as the sled slows to a stop.
Conversion of Units Made Simple
When solving physics problems, it's often necessary to convert units to make calculations easier. In most physics problems, the standard unit for speed is meters per second (m/s), while the given speed may initially be in kilometers per hour (km/h). Here’s how we make that conversion.
The conversion factor from kilometers per hour to meters per second is that 1 km/h equals \(\frac{1}{3.6}\) m/s. This means:
  • To convert km/h to m/s, multiply the speed by \(\frac{1}{3.6}\).
  • In our exercise, \(1020 \text{ km/h} \times \frac{1}{3.6} = 283.33 \text{ m/s}\).
Knowing how to convert units is crucial for accuracy in physics as it ensures that the numbers you work with make sense in the context of the problem. It's like changing languages; different units tell what kind of "language" you need to speak for that particular part of the calculation.
Grasping Deceleration in Physics
Deceleration is simply acceleration in the opposite direction of motion, often referred to as negative acceleration. It’s what happens when an object slows down over time, and in the context of our exercise, it's what happened to Colonel Stapp's sled as it came to a stop.
When calculating deceleration, we still use the standard acceleration formula \(a = \frac{\Delta v}{\Delta t}\), but this time, \(\Delta v\) represents a decrease in velocity. The final result is thus a negative value indicating slowing down.
In terms of physics:
  • Deceleration means the final velocity \(v_f\) is less than the initial velocity \(v_i\), and the velocity change \(\Delta v = v_f - v_i\) becomes negative.
  • For Colonel Stapp's ride, the initial speed was 283.33 m/s, dropping to 0 m/s, so the change in velocity is negative.
Expressing the deceleration in terms of gravity (\(g\)), which is approximately \(9.81 \text{ m/s}^2\), makes it easier to comprehend how strong the force was - up to about \(20.63\) times stronger than regular gravitational pull! This helps visualize the immense force involved in bringing that speeding sled to a stop.

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Most popular questions from this chapter

A stone is dropped into a river from a bridge \(43.9 \mathrm{~m}\) above the water. Another stone is thrown vertically down \(1.00 \mathrm{~s}\) after the first is dropped. The stones strike the water at the same timc. (a) What is the initial speed of the scoond stonc? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released.

The position of a particle moving along an \(x\) axis is given by \(x=12 t^{2}-2 t^{3}\), where \(x\) is in meters and \(t\) is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at \(t=3.0 \mathrm{~s}\). (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and \(\underline{(g)}\) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at \(t=0\) )? (i) Determine the average velocity of the particle between \(t=0\) and \(t=\overline{3} \mathrm{~s}\)

(a) If a particlc's position is given by \(x=4-12 t+3 t^{2}\) (where \(t\) is in seconds and \(x\) is in meters), what is its velocity at \(t=1 \mathrm{~s} ?(\mathrm{~b})\) Is it moving in the positive or negative direction of \(x\) just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (c) Is there cver an instant when the velocity is zero? If so. give the time \(r\), if not, answer no. (f) Is there a time after \(t=3 \mathrm{~s}\) when the particle is moving in the ncgative dircction of \(x ?\) If so, give the time \(t\), if not, answer no.

A red train traveling at \(72 \mathrm{~km} / \mathrm{h}\) and a green train traveling at \(144 \mathrm{~km} / \mathrm{h}\) are headed toward each other along a straight, level track. When they are \(950 \mathrm{~m}\) apart, each engineer sees the other's train and applies the brakes. The brakes slow each train at the rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2} .\) Is there a collision? If so, answer yes and give the speed of the red train and the speed of the green train at impact, respectively. If not, answer no and give the separation between the trains when they stop.

You drive on Interstate 10 from San Antonio to Houston, half the time at \(55 \mathrm{~km} / \mathrm{h}\) and the other half at \(90 \mathrm{~km} / \mathrm{h}\). On the way back you travel half the distance at \(55 \mathrm{~km} / \mathrm{h}\) and the other half at \(90 \mathrm{~km} / \mathrm{h}\). What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip? (e) Sketch \(x\) versus \(t\) for (a), assuming the motion is all in the positive \(x\) direction. Indicate how the average velocity can be found on the sketch.

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